Question

Find the smallest number that must be multiplied to 768 to make it perfect square also find the square root of the number this obtained

Answer 1

i) 252 The prime factorization of 252 is 252=2×2×3×3×7. As the prime factor 7 has no pair, 252 is not a perfect square. If 7 gets a pair, then the number will be a perfect square. So, we multiply 252 by 7 to get

2 252
2 126
3 63
3 21
7
252×7=2×2−−−−×3×3−−−−×7×7−−−− Now each prime factor has a pair. Therefore, 252×7=1764 is a perfect square. Thus the required smallest number is 7. Thus, 1764−−−−√=2×3×7=42. (ii) 180 The prime factorization of 180 is180=2×2×3×3×5. As the prime factor 5 has no pair, 180 is not a perfect square. If 5 gets a pair, then the number will be a perfect square. So, we multiply 180 by 5 to get
2 180
2 90
3 45
3 15
5
180×5=2×2−−−−×3×3−−−−×5×5−−−−. Now each prime factor has a pair. Therefore, 180×5=900 is a perfect square. Thus the required smallest number is 5. Thus, 900−−−√=2×3×5=30 (iii) 1008 The prime factorization of 1008 is1008=2×2×2×2×3×3×7. As the prime factor 7 has no pair, 1008 is not a perfect square. If 7 gets a pair, then the number will be a perfect square. So, we multiply 1008 by 7 to get
2 1008
2 504
2 252
2 126
3 63
3 21
7
1008×7=2×2−−−−×2×2−−−−×3×3−−−−×7×7−−−− Now each prime factor has a pair. Therefore, 1008 x 7 = 7056 is a perfect square. Thus the required smallest number is 7. Thus,7056−−−−√=2×2×3×7=84. (iv) 2028 The prime